# If sum of infinite terms of a G.P. is 3 and sum of squares of these terms is also 3 then its first and common ratio is a) 1,1/2 b)3/2,12 c) 1/2,3/2 d) none If roots of equation a(b-c)x^2 + b(c-a)x + c(a-b)=0 are equal than a,b,c are in a) A.P. B)G.P. C) H.P. D)NONE plzzzzz solve these question

1
by kaushikravikant

2015-06-14T01:09:00+05:30

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A,ar,ar^2,ar^3,.....infinite number of terms

Here r < 1.  So the sum converges.

Sum = a / (1 - r) = 3        =>  a = 3 (1 - r)   ---- (1)

Series with terms being squares of the above series:
a², a² r², a²r⁴, a² r⁶, a² r⁸ ,...... infinity..

Sum of the terms = a² / ( 1 - r²)  = 3
=>    a² = 3 (1 - r²)    --- (2)

divide (2) by (1):      a = 1 + r    --- (3)

(1) - (3) :    2 - 4 r = 0        =>  r = 1/2
hence,  by (3) ,   a = 3/2
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If the Roots of the equation a (b-c) x² + b (c-a) x + c (a-b) = 0  are equal it means that the discriminant is 0.

Discriminant  =   b² (c-a)² - 4*a (b-c) * c (a-b)  = 0  --(1)

If a, b, c are in AP,  then:   b - a = c - b    and  c - a = 2 (c - b)  = 2 (b - a)
Substituting in (1),     b² 2² (c - b)² = 4 a c (b - c)(a - b)
=>  b² (c - b) = a c (a - b)

If a, b, c are in GP:  b² = a c
=>    (c - a)² = 4 (b - c) (a - b)
=>      c² + a² + 2 ac = 4 a b - 4 b² + 4 b c
=>   (c + a)²  =  4 b (a + c) - 4 b²

If a,b,c are in HP then:  1/a, 1/b , 1/c are in AP.
=>  1/b - 1/a = 1/c - 1/b
=>  c (a - b)  = a (b - c)
Substituting in (1):
b² (c -a)² = 4 a² (b - c)²
=>  b ( c - a) = + 2 a (b - c)  = + 2 c (a - b)     --- (2)
OR  b (c - a) = - 2 a (b - c)  = - 2 c (a - b)  --- (3)

By (2)      =>  (b + 2 a) c =  3 a b      and  3 b c = a (b +2 c)
=>  9 a b - 6 a c = a b + 2 a c
=>  8 a  b = 8 a c         =>  b = c      or    a = 0

By (3):    =>    a b + 2 a c + b c = 0      =>  1/c + 2/b + 1/a = 0
and  b c + a b - 2 c a = 0      =>  1/a + 1/c - 2/b = 0

Hence,  a,b,c are not in AP, GP or HP.

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